Q: What does IP Maths Notes (Lower Sec, Year 1-2): 08) Trigonometry in Right Triangles cover? A: Apply sine, cosine, and tangent to height-distance problems, bearings, and angles of elevation with calculator and non-calculator techniques.
The core idea is simple: Label opposite, adjacent, and hypotenuse before choosing a trig ratio.
Use it as a working check: Use sine, cosine, or tangent when the triangle is right-angled. Use inverse trig when the angle is unknown, and always draw elevation, depression, or bearing diagrams first.
Then go one layer deeper: Work through the side, angle, drone, and bearing examples to practise translating a word problem into a right triangle before calculating.
Trigonometry unlocks height and distance tasks across science and geography contexts. Become fluent with ratios, inverse functions, and diagram interpretation.
These notes align with MOE Lower Secondary Mathematics syllabus used in IP pathways (aligned to O-Level Mathematics 4052 foundations).
Status: MOE Lower Secondary Mathematics syllabus (latest release) checked 2025-11-30 - scope unchanged; remains the reference for these lower-sec notes.
Opposite and adjacent depend on this angle, not on the diagram's orientation.
3
The side directly across from θ.
This is the opposite side.
4
The side touching θ that is not the hypotenuse.
This is the adjacent side.
Then choose the ratio from the two sides mentioned in the question:
Known and unknown sides
Ratio to start with
Opposite and hypotenuse
sinθ=hypotenuseopposite
Adjacent and hypotenuse
cosθ=hypotenuseadjacent
Opposite and adjacent
tanθ=adjacentopposite
Worked example - Finding a side
Given right triangle with angle θ=37∘ and hypotenuse 12 cm, find the opposite side.
sin37∘=12opp⟹opp=12sin37∘≈12×0.601=7.21cm.
Worked example - Finding an angle
Opposite side 5 cm, adjacent side 8 cm. Find θ.
tanθ=85⟹θ=tan−1(85)≈32.0∘.
Worked example - Choosing the ratio from labels
A ramp makes an angle of 18∘ with the ground. The horizontal distance along the ground is 6.0m. Find the vertical height gained.
The angle is measured from the ground, so the ground distance is adjacent.
The vertical height is opposite the angle.
Opposite and adjacent point to tangent:
tan18∘=6.0h
h=6.0tan18∘≈1.95m.
Misconception check: do not use sine just because the question asks for a height. Sine is only for opposite and hypotenuse. Here the given length is adjacent, so tangent is the correct ratio.
2 Exact values and surds
Memorise sin30∘=21, cos45∘=22, tan60∘=3.
Use rationalisation to present answers cleanly: if sinθ=23, then θ=60∘ or other equivalent angles depending on context.
Exact-value triangle checkpoint
Use exact values as triangle ratios, not as isolated facts. Start from the special triangle, then pick opposite, adjacent, and hypotenuse relative to the angle named in the question.
Triangle
Side ratio to remember
If the focus angle is...
Example ratio
Common trap
30-60-90
short : long : hypotenuse = 1:3:2
30∘
sin30∘=21
Using the side opposite 60∘ by mistake.
30-60-90
short : long : hypotenuse = 1:3:2
60∘
45-45-90
equal : equal : hypotenuse = 1:1:2
45∘
Worked check: if cos30∘ is needed, use the 30-60-90 triangle and look at the side adjacent to 30∘. The adjacent side is 3 and the hypotenuse is 2, so cos30∘=23.
Misconception check: the special triangle does not decide the ratio for you. The ratio still comes from SOH-CAH-TOA after you identify the focus angle.
3 Angle of elevation/depression
Angle of elevation: measured upwards from horizontal.
Angle of depression: measured downwards from observer to object.
Before calculating, translate the words into a right triangle:
Phrase in the question
Draw this first
Common trap
Angle of elevation
Horizontal eye-level line, then line of sight upwards.
Measuring the angle from the vertical.
Angle of depression
Horizontal eye-level line, then line of sight downwards.
Forgetting it equals the alternate interior angle at ground level when horizontals are parallel.
Bearing
North line at the starting point, then clockwise angle to the path.
Treating bearing as an ordinary angle inside the triangle before drawing north.
Height or depth
Vertical side of the right triangle.
Using sine automatically instead of checking which length is given.
Horizontal distance
Ground line or parallel horizontal line.
Confusing it with the slanted line of sight.
Only choose sine, cosine, or tangent after the right angle, focus angle, and two relevant sides are labelled.
Worked example - Two-point observation
A drone lifts off and reaches a point where the angle of elevation from a student 50 m away is 35∘. What is the drone height?
tan35∘=50h⟹h=50tan35∘≈35.0m.
Two-observer height checkpoint
When two observers on level ground look at the same vertical object, draw one baseline and split it into two horizontal distances. Use the same height in both tangent equations.
Step
What to define
Why it helps
1
Let the unknown distance from the nearer observer to the base be x.
The farther observer's distance can be written using the given separation.
2
Write the other horizontal distance as x+d or d−x, depending on the diagram.
This keeps both right triangles tied to the same baseline.
3
Use tan(angle)=horizontal distanceheight for each observer.
Both equations share the same vertical height.
4
Equate the two height expressions, then solve for x before finding the height.
Solving the distance first avoids guessing which triangle gives the answer.
Worked check: if two observers are 40m apart and the nearer angle is larger, the nearer horizontal distance should be the smaller one. If your algebra makes the larger angle use the larger distance, the diagram labels are probably reversed.
Misconception check: do not add the two angles of elevation. They belong to two different right triangles that share a height, not one triangle with two base angles.
4 Bearings and composite paths
Bearings measured clockwise from north. Break vectors into horizontal/vertical components using cosine and sine, then recombine with Pythagoras.
Bearing component checkpoint
For bearing questions, draw a north line first, then split each journey into east-west and north-south components. Treat north and east as positive unless the question gives a different convention.
Bearing type
East-west component
North-south component
Common trap
θ east of north, such as 040∘
+dsinθ
+dcosθ
Swapping sine and cosine because the angle is drawn near the horizontal.
θ west of north, such as 320∘
−dsin40∘
+dcos40∘
θ east of south, such as 140∘
+dsin40∘
−dcos40∘
Two-leg journey
Add all east-west components, then add all north-south components.
Use Pythagoras only after components are combined.
Finding each leg's displacement and adding lengths directly.
Worked check: a walk of 10km on a bearing of 060∘ has east component 10sin60∘=8.66km and north component 10cos60∘=5.00km. The sine term is east because the bearing angle is measured from north, not from the horizontal.
Preview: sine and cosine rules
Although formally covered in upper sec, introduce the formulas for awareness:
sinAa=sinBb=sinCc,c2=a2+b2−2abcosC.
Practice Quiz
Check SOH-CAH-TOA recognition, diagram labelling, elevation problems, and bearings decomposition with the quiz below.
Try it yourself
Find the hypotenuse of a right triangle with adjacent side 9 cm and angle 28∘.
A ladder leans against a wall with its base 1.6 m from the wall and top 3.8 m above ground. Find the ladder length and angle of elevation.
Expected line length:1.62+3.82≈4.1m (angle of elevation ≈67.6∘).
Two observers stand 40 m apart on level ground. They observe the top of a pole with angles of elevation 32∘ and 48∘ respectively. Model the scenario and compute the pole height.
A yacht sails 12 km on a bearing of 040∘ then 8 km on 130∘. Calculate displacement from the starting point.
